Since you have no assumption on $K$ (except compactness), I don't see the connection with (twisted) Gelfand pairs... In particular, taking for $K$ the identity subgroup, you get the convolution algebra of compactly supported functions on $G$. Is this really what you want?
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Alain ValetteJan 22 '12 at 22:30

Usually one wants this algebra to be commutative to be a twisted Gelfand pair.
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Benjamin SteinbergJan 22 '12 at 23:57

Of course, I am mostly interested in some non trivial compact subgroups such as the maximal compact in the case of $GL(N)$. Commutativity doesn't hold in general, I guess?
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Marc PalmJan 23 '12 at 7:51

My understanding is that the term twisted Gelfand pair is only used when the algebra is commutative.
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Benjamin SteinbergJan 26 '12 at 14:26

2 Answers
2

These Hecke algebras are intensively studied in the field of "type theory" for reductive $p$-adic groups.

You have a nice summary of basic facts with proofs in chapter 4 of Bushnell and Kutzko's book "The admissible dual of ${\rm GL}(N)$ via compact open subgroups" (the chapter is entitled "Interlude with Hecke algebras").

You may also read the monography "The Langlands conjecture for ${\rm GL}(2)$", written by Bushnell and Henniart. You'll find there a nice introduction to these algebras.

There are many other references. But it depends on what exactly you're interested in.

A classical reference for twisted Gelfand pairs is J.R. Stembridge, On Schur's Q-functions and the primitive idempotents of a commutative Hecke algebra, J. Algebr. Comb. 1 (1992) 71–95, but I believe this paper considers only finite groups.